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3 years ago

Rewrite 2x2x2 using an exponent.

Mathematics

1 answer:

Nastasia [14]3 years ago

6 0

Answer:

2³

Step-by-step explanation:

please mark me as brainliest

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Bill can type 19 words per minute faster than bob. their combined typing speed is 97 words per minute. find bobs typing speed

Marina86 [1]

Answer:

Bob's typing speed is 39 words per minute.

Bill's typing speed is 58 words per minute.

Step-by-step explanation:

Given : 

Bill can type 19 words per minute faster than bob. their combined typing speed is 97 words per minute.

To Find : 

Bobs typing speed

Solution : 

Let the bobs typing speed is x

According to the given question :

Bobs typing speed is x

Bill's typing speed is 19+x

Their combined speed is 97 words per minute.

Solving value x : 

x + 19+x = 97

2x + 19 = 97

2x = 97 - 19

2x = 78

x = 78/2

x = 39

Bobs typing speed = 39

Bill's typing speed 39 + 19 = 58

Hence, Bobs speed is 39 words per minute.

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3 years ago

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WILL GIVE BRAINLIEST!! PLEASE HELP ASAP!! (Include question number)

alexgriva [62]

Answer:

The first one is F' each one is around $36

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3 years ago

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35/50 into percentage

lara [203]

35/50 as a percentage is 70%!

6 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

H(t) = -10+ - 6 h()= -44

lesya [120]

Answer:

-44

Step-by-step explanation:

4 0

3 years ago